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$R$ ring and $M$ a left $R$-module. Call $\mathrm{Soc}\;M$ the sum of all the simple submodules of $M$. Then

$M$ is artinian if and only if $\lambda_R(\mathrm{Soc}(M))<\infty$ and for very $0\neq Q\subset M$ we have $Q\cap\mathrm{Soc}\;M\neq0$

Could you help me solve this exercise?

($\lambda_R$ denotes the composition length)

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Is this homework? Are you conjecturing this is true? You picked the statement from a book? Have you tried any approach? Can you do at least one of the two implications? (You've asked almost a 100 questions by now!) Probably telling us what $\lambda_R$ means would also be a good idea :D – Mariano Suárez-Alvarez May 18 '12 at 2:49
I'm doing exercises by myself, I found this in an old exam. $\lambda_R(M)$ is the length of $M$ as an $R$-module – Alex M May 18 '12 at 3:38
@AlexM Please define "length" specifically. There is more than one notion of length. The statement is false for the first two notions that came to my mind. It obviously can't mean "composition length" because there are Artinian modules over commutative rings with infinite composition length. It also can't be uniform dimension because there exists a commutative non-Artinian uniserial ring with a minimum ideal which serves as a counterexample. – rschwieb May 18 '12 at 19:24
you're right, it should be $\lambda_R(\mathrm{Soc}(M))$ and the notion is the "composition length" – Alex M May 18 '12 at 23:52
Please add all this information to the question itself, so that people do not have to read all the comments! :D – Mariano Suárez-Alvarez May 18 '12 at 23:55

If $M$ is artinian, its socle $\newcommand\soc{\operatorname{soc}}\soc M$, which is a submodule, is also artinian. Since $\soc M$ is semisimple, being artinian it is also noetherian, and therefore has finite length. Moreover, if $Q$ is a non-zero submodule of $M$, then it is also artinian, and has a non-zero socle: to check this, one has to show that $Q$ contains a simple module, but if it didn't we could easily construct a non-stopping decreasing sequence of submodules. Finally, $\soc Q$, being a non-zero sum of simple submodules of $Q$ (so of $M$) intersects $\soc M$ non-trivially.

Can you do the converse using similar ideas?

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I took a descending chain $\ldots\subset M_2\subset M_1\subset M$ with $M_i\neq 0$, then I intersecated with $\mathrm{Soc}\;M$, and since it is artinian we have for some $n$ that $M_n\cap\mathrm{Soc}\;M=M_{n+1}\cap\mathrm{Soc};M$, I want to prove that $M_n\subset M_{n+1}$. I proved that every simple submodule of $M_n$ is contained in $M_{n+1}$ but I don't know how to continue, any help? – Alex M May 28 '12 at 7:07

With the clarification made that we are talking about composition length, and that the socle should have finite length, the proposed statement is false.

A module is called finitely cogenerated if it has a finitely generated essential socle. (Here the "finitely generated" may be swapped for "finite composition length", since the two notions are identical for a semisimple module.)

So, the question amounts to asking "Show a module is Artinian iff it is finitely cogenerated." From Mariano's answer, we know that Artinian modules are finitely cogenerated, but the converse is false. A correct statement would be that "a module $M$ is Artinian iff $M/N$ is finitely cogenerated for every submodule $N$ of $M$."

There exists a non-Artinian commutative ring $R$ such that $R_R$ is a finitely cogenerated module. The first example I located is due to Osofsky and can be found in Lam's Lectures on Modules and Rings at the top of page 514.

If you request, I can duplicate it here.

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